Continued Fractions

We'll approach continued fractions from a completely different viewpoint.

Let $A_1,...$ be a sequence of $2\times 2$ matrices. Let

$$\left(\begin{array}{cc}{p}_n& r_n\\ q_n& s_n\end{array}\right)=A_1...A_n$$

If $\frac{p_n}{q_n}$ and $\frac{r_n}{s_n}$ tend to the same limit $\alpha$ as $n\to \mathrm{\infty }$ then $\alpha$ is the limit of the infinite product of matrices, and we write

$$\alpha ~\prod _{n=1}^{\mathrm{\infty }}{A}_n$$

Example: Let $a_1,...$ be a sequence of positive integers, except we allow $a_1=0$. Let

$$A_k=\left(\begin{array}{cc}{a}_k& 1\\ 1& 0\end{array}\right)$$

Then $[a_1;a_2,...]~{\prod }_{n=0}^{\mathrm{\infty }}{A}_n$. The convergents are $p_n,q_n$, and we have $r_n=p_{n-1},s_n=q_{n-1}$. Since $\det A_n=-1$, we see the difference between successive convergents can be determined by studying

$$\det \mid \begin{array}{cc}{p}_n& r_n\\ q_n& s_n\end{array}\mid =(-1{)}^n$$

Example: Suppose $\alpha ~{\prod }_{k=1}^{\mathrm{\infty }}{A}_k$.

Let $c_k$ be nonzero numbers. Then $\alpha ~{\prod }_{k=1}^{\mathrm{\infty }}{c}_k{A}_k$.

Let $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Then

$$\frac{a\alpha +b}{c\alpha +d}~A\prod _{k=1}^{\mathrm{\infty }}{A}_k$$

provided $c\alpha +d\ne 0$. For some proofs, we exploit this and ignore a finite number of matrices.

Example: Given that $\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)...$ converges, find its limit.

Let

$$\alpha ~\prod _{n=1}^{\mathrm{\infty }}\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)$$

Then

$$\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\prod _{n=2}^{\mathrm{\infty }}\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)~\frac{\alpha +2}{\alpha +1}$$

Hence $\alpha =\frac{\alpha +2}{\alpha +1}$, and therefore $\alpha =\sqrt{2}$.

Theorem: Consider a sequence of matrices of the form $A_k=\left(\begin{array}{cc}{a}_k& c_k\\ b_k& 0\end{array}\right)$. where $a_k,b_k,c_k$ are positive integers satisfying

$$a_k\ge (b_k{c}_k{)}^{1+\delta }$$

for some fixed $\delta >0$. Then ${\prod }_{n=1}^{\mathrm{\infty }}~\alpha$ for some irrational $\alpha >1$.

Proof: Let ${\prod }_{k=1}^n{A}_k=\left(\begin{array}{cc}{p}_n& r_n\\ q_n& s_n\end{array}\right)$. We have $r_n=c_n{p}_{n-1},s_n=c_n{q}_{n-1}$, $q\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$ and $p_n\ge q_n$. The last fact implies $\alpha >1$.

Also,

$$\frac{p_{n+1}}{q_{n+1}}=\frac{a_{n+1}{p}_n+b_{n+1}{c}_n{p}_{n-1}}{a_{n+1}{q}_n+b_{n+1}{c}_n{q}_{n-1}}$$

so $\frac{p_{n+1}}{q_{n+1}}$ lies strictly between $\frac{p_n}{q_n}$ and $\frac{p_{n-1}}{q_{n-1}}$.

Then:

$$\mid \frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}\mid =\frac{\prod _{k=1}^n\mid \det A_k\mid }{c_n{q}_n{q}_{n-1}}=\frac{b_1{c}_1...b_n{c}_n}{c_n{q}_n{q}_{n-1}}<\frac{b_1{c}_1...b_{n-1}{c}_{n-1}}{(q_{n-1}{)}^2}$$

since $q_n>b_n{q}_{n-1}$

Now $q_{n-1}>(k{b}_1{c}_1...b_{n-1}{c}_{n-1}{)}^{1+\delta }$ for some $\kappa >0$ because

$$\begin{array}{rlr}{q}_{n-1}& =& a_{n-1}{q}_{n-2}+b_{n-1}{c}_{n-2}{q}_{n-3}\\ & \ge & a_{n-1}{q}_{n-2}\ge (b_{n-1}{c}_{n-1}{)}^{1+\delta }{q}_{n-2}\end{array}$$

and if we asssume inductively that $q_{n-2}>(\kappa b_1{c}_1...b_{n-2}{c}_{n-2}{)}^{1+\delta }$ it follows that

$$q_{n-1}>(\kappa b_1{c}_1...b_{n-1}{c}_{n-1}{)}^{1+\delta }$$

So $\mid \frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}\mid \to 0$ as $n\to \mathrm{\infty }$ and

$$\mid \alpha -\frac{p_{n-1}}{q_{n-1}}\mid \le \mid \frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}\mid <\frac{b_1{c}_1...b_{n-1}{c}_{n-1}}{q_{n-1}^2}<\frac{1}{\kappa q_{n-1}^{1+\sigma }}$$

where $\sigma >0$ is defined by $\frac{1}{1+\delta }=1-\sigma$.

Hence $\alpha$ is irrational, because if $\alpha =k/l$ then:

$$\mid k{q}_n-l{p}_n\mid =l{q}_n\mid \alpha -\frac{p_n}{q_n}\mid <\frac{l{q}_n}{\kappa q_n^{1+\sigma }}\to 0$$

as $n\to \mathrm{\infty }$, a contradiction because $\mid k{q}_n-l{p}_n\mid$ is always a nonzero integer (since $\frac{p_{n+1}}{q_{n+1}}$ lies strictly between the previous two convergents).

Corollary: Let $a_k$ be positive integers. Then $[a_1;a_2,....]$ converges to an irrational $\alpha >1$.

Proof: Apply theorem to continued fraction matrices.

Example: ${\prod }_{k=1}^{\mathrm{\infty }}\left(\begin{array}{cc}k& 1\\ 2& 0\end{array}\right)$ converges to an irrational number because apart from the first two matrices, we have

$$a_k=k>2^{1.5}=(b_k{c}_k{)}^{1.5}$$

We can develop this theory further to derive continued fraction expansions of hyperbolic trignometric functions.